1. We need to identify the process for which entropy change, $\Delta S$, is negative.

Entropy generally:

• increases when randomness/disorder increases,
• decreases when randomness/disorder decreases.

Now check each option.

2. Option A: $H_2(g) \to 2H(g)$

Here, one gaseous molecule breaks into two gaseous atoms.

• Number of particles increases.
• Disorder increases.

Therefore, $\Delta S > 0$ So, this is not the correct option.

3. Option B: $N_2(g, 1\,\text{atm}) \to N_2(g, 5\,\text{atm})$

This is compression of a gas from lower pressure to higher pressure.

• Gas becomes more ordered.
• Available volume effectively decreases.
• Entropy decreases.

For an ideal gas at constant temperature, $\Delta S = nR \ln\left(\frac{V_2}{V_1}\right) = -nR \ln\left(\frac{P_2}{P_1}\right)$ Since $P_2 > P_1$, $\ln\left(\frac{P_2}{P_1}\right) > 0 \implies \Delta S < 0$

Hence, Option B gives negative entropy change.

4. Option C: $C(\text{diamond}) \to C(\text{graphite})$

Graphite has greater entropy than diamond because graphite has a more stable layered structure with greater disorder/freedom compared to diamond.

Thus, $S_{\text{graphite}} > S_{\text{diamond}}$ So, $\Delta S > 0$ This is not correct.

5. Option D: $N_2(g, 273\,K) \to N_2(g, 300\,K)$

Heating a gas increases molecular motion and disorder. Therefore, $\Delta S > 0$ So, this is not the correct option.

6. Conclusion

Only Option B has negative entropy change.

$\boxed{\Delta S < 0 \text{ for option B}}$